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DedPeter [7]
3 years ago
5

Two children are pulling on opposite sides of a blanket. The brother is pulling with a force of 3 N. The sister is pulling with

a force of 5 N. What is the net force? In which direction would the blanket move: toward the brother or toward the sister?
Physics
1 answer:
Citrus2011 [14]3 years ago
7 0
Let F1=Force exerted by the brother (+F1)
F1= Force exerted by the sister (-F2)

Fnet=(+F1) + (-F2)
Fnet= (+F1) + (-F2)
Fnet=F1 - F2
Fnet= (+3N)+(-5N)
Fnet= -2N

-F

towards the sister (-F) (greater force applied)
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A force does 30000 J of work along a distance of 9.5m. Find the applied force.
Elina [12.6K]

Answer: F=3158N

Explanation:

Work is the product of force applied and the distance the object moves along the force applied. Work is measured in joules and the equation is as follows

W = F x d

So that F = W/d

F = 30000 J / 9.5m

F = ~3158 N

8 0
3 years ago
With what minimum speed must you toss a 130 gg ball straight up to just touch the 15-mm-high roof of the gymnasium if you releas
xxTIMURxx [149]

Answer:

The initial velocity is 0.5114 m/s or 511.4 mm/s

Explanation:

Let the initial velocity be 'v'.

Given:

Mass of the ball (m) = 130 g = 0.130 kg   [ 1 g = 0.001 kg]

Initial height of the ball (h₁) = 1.4 mm = 0.0014 m   [ 1 mm = 0.001 m]

Final height of the ball (h₂) = 15 mm = 0.015 m

Now, from conservation of energy principle, energy can neither be created nor be destroyed but converted from one form to another.

Here, the kinetic energy of the ball is converted to gravitational potential energy of the ball after reaching the final height.

Change in kinetic energy is given as:

\Delta KE=\frac{1}{2}m(v_f^2-v_i^2)\\Where\ v_f\to Final\ velocity\\v_i\to Initial\ velocity

As it just touches the 15 mm high roof, the final velocity will be zero. So,

v_f=0\ m/s.

Now, the change in kinetic energy is equal to:

\Delta KE = \frac{1}{2}\times 0.130\times v^2\\\\\Delta KE = 0.065v^2

Change in gravitational potential energy = Final PE - Initial PE

So,

\Delta U=mg(h_f-h_i)\\\\\Delta U=0.130\times 9.8\times (0.015-0.0014)\\\\\Delta U=0.017\ J                    [ g = 9.8 m/s²]

Now, Change in KE = Change in PE

0.065v^2=0.017\\\\v=\sqrt{\frac{0.017}{0.065}}\\\\v=0.5114\ m/s\\\\1\ m=1000\ mm\\\\So,0.5114\ m=511.4\ mm\\\\\therefore v=511.4\ mm/s

Therefore, the initial velocity is 0.5114 m/s or 511.4 mm/s

4 0
3 years ago
The drag force pushes opposite your motion as you ride a bicycle. If you double your speed, what happens to the drag force?
timurjin [86]

Answer: The drag force goes up by a factor of 4

Explanation:

The <u>Drag Force</u> equation is:

F_{D}=\frac{1}{2}C_{D}\rho A_{D}V^{2} (1)

Where:

F_{D} is the Drag Force

C_{D} is the Drag coefficient, which depends on the material

\rho is the density of the fluid where the bicycle is moving (<u>air in this case) </u>

A_{D} is the transversal area of the body or object

V the bicycle's velocity

Now, if we assume C_{D}, \rho and A_{D} do not change, we can rewrite (1) as:

F_{D}=C.V^{2} (2)

Where C groups all these coefficients.

So, if we have a new velocity V_{n} , which is the double of the former velocity:

V_{n}=2V (3)

Equation (2) is written as:

F_{D}=C.V_{n}^{2}=C.(2V)^{2}

F_{D}=4CV^{2} (4)

Comparing (2) and (4) we can conclude<u> the Drag force is four times greater when the speed is doubled.</u>

7 0
3 years ago
Enny and Anne wanted to see who could throw a ball the hardest. They decided to each throw a ball against a wall as hard as they
mihalych1998 [28]

Answer:

wha t kind of variables

independent variables

dependent variables

controled variables

4 0
2 years ago
Mathphys :( im sorry i annoy you
Vitek1552 [10]

Answer:

4. 7.59276

Explanation:

Add up the x components:

Aₓ + Bₓ + Cₓ = 5 − 1.6 + 2.4 = 5.8

Add up the y components:

Aᵧ + Bᵧ + Cᵧ = -2.4 + 3.3 + 4 = 4.9

Use Pythagorean theorem to find the magnitude:

√(x² + y²)

√(5.8² + 4.9²)

√57.65

7.59276

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3 years ago
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